In some cases, people work with quasi-symmetric maps of $\mathbb{R}$ instead of $\mathbb{S}^1$. More precisely, we say that an increasing homeomorphism $\phi \colon \mathbb{R} \to \mathbb{R}$ is quasi-symmetric if there exists $M > 0$ such that
$$ 1/M \leq \frac{|\phi(x + t) - \phi(x)|}{|\phi(x - t) - \phi(x)|} \leq M, $$
where $x \in \mathbb{R}$ and $t > 0$ are arbitrary.
If we have such a map $\phi\colon \mathbb{R} \to \mathbb{R}$, then using some Mobius map $f$ sending the upper half-plane $\mathbb{H}$ to the unit disk $\mathbb{D}$ centered at zero, we can obtain an orientation-preserving homeomorphism $\tilde{\phi} \colon \mathbb{S}^1 = \partial \mathbb{D} \to \mathbb{S}^1$, which is defined as $\tilde{\phi}(x) = (f \circ \phi \circ f^{-1})(x)$ if $x \in f(\mathbb{R})$, and by continuity for the point $x_0 = f(\infty)$.
However, even though $\phi$ is quasi-symmetric on $\mathbb{R}$, then $\tilde{\phi}$ might not be quasi-symmetric on $\mathbb{S}^1$. For instance, piecewise quasi-symmetric maps on $\mathbb{S}^1$ might not be quasi-symmetric.
My question is the following, which conditions should we put on $\phi$ defined on $\mathbb{R}$ so that the corresponding homeomorphism $\tilde{\phi}$ defined on $\mathbb{S}^1$ will be quasi-symmetric?
In fact, I am interested in some particular example of a quasi-symmetric on $\mathbb{R}$ map. In Theorem 6 in the paper Boundary correspondence under Quasiconformal mappings by J.A. Kelingos (https://projecteuclid.org/journals/michigan-mathematical-journal/volume-13/issue-2/Boundary-correspondence-under-quasiconformal-mappings/10.1307/mmj/1028999549.full), he shows that quasi-symmetric map $u \colon [0, 1] \to [0, 1]$ can be extended to the whole real line as a quasi-symmetric and continuously differentiable outside of $[0,1]$ map. I am interested in his construction provides a quasi-symmetric map with analogous properties in the case of $\mathbb{S}^1$.